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11_danielbrito.cpp
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11_danielbrito.cpp
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// https://neps.academy/problem/11
// First of all, let's analyze some patterns in the problem, taking as an example this case:
// 3 -> N value
// 3 2 1 -> Possbile permutation
// 2 1 3 -> Possbile permutation
// 1 3 2 -> Possbile permutation
// 3 1 2 -> Possbile permutation
// 1 2 3 -> Possbile permutation
// 10 9 11 -> Adding up the columns of the given numbers (subtotal)
// 2 3 1 -> Missing permutation (Notice that is the difference between 'total'-'subtotal'.
// 12 12 12 -> Sum of all the values (total)
// So, we just need to figure out some formula to calculate the 'total' and then subtract the 'subtotal' from it.
// In the end, this operation will give us the missing permutation.
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
int main(){
int n, i, j, x, fat=1;
cin >> n;
// Calculating the factorial of 'n'.
// It'll give us the total number of permutations.
for(i=1; i<=n; i++) fat *= i;
// This vector will be used to store the sum of the values of each column:
vector<int> subtotal(n, 0);
// Notice that we use 'fat-1' because there is a missing permutation.
for(i=0; i<fat-1; i++){
for (j=0; j<n; j++){
cin >> x;
// Notice that we don't need to store the hole matrix, just the sum of the values of each 'column':
subtotal[j] += x;
}
}
// This vector will be used to store the numbers of the missing permutation, as explained before:
vector<int> res;
// Here, we use the concept of Arithmetic Progression.
// This will help us to calulate the 'total' sum of a column with all the possible permutations:
int apSum = ((1+n)*n)/2;
// This loop is used to traverse 'subtotal' and subtract its value from 'total', to calculate their difference:
for(i=0; i<n; i++){
// I struggled to figure out how to find the sum of the values of a column as if it has all the permutations (total).
// Finally, I ended up with this formula: apSum*(fat/n). So, we just calculate the difference.
res.push_back(apSum*(fat/n)-subtotal[i]);
}
// Printint out the result, that is the missing permutation:
for(int i=0; i<res.size(); i++){
// Notice that we can't print out a space after the last number of the permutation.
// Otherwise, it should give us a 'Presentation error'.
(i<n-1) ? cout << res[i] << " " : cout << res[i];
}
cout << "\n";
return 0;
}